91Ó°ÊÓ

The electric force experienced by a charged particle in an electric field relies on the particle's charge and the electric field's strength.

The expression for the force is given as:

\(F = QE\)

Here, Q is the charge and E is the electric field strength.

The electric field strength is, \(E = 150\;{\rm{N/C}}\)

The radius of the droplet is, \(r = 0.018\;{\rm{mm}}\).

The mass of droplet is given by,

\(m = \rho V\)

Here, \(\rho \)is the density of water and V is the volume of the droplet.

In equilibrium conditions, the electric force experienced by the droplet is equal to the weight of the droplet.

\(\begin{aligned}{c}F &= nqE\\mg &= nqE\\\rho \left( {\frac{4}{3}\pi {r^3}} \right)g &= nqE\\n &= \frac{{\rho \left( {\frac{4}{3}\pi {r^3}} \right)g}}{{qE}}\end{aligned}\)

Here, nis the number of electrons, qis the charge of electrons and g is the gravitational acceleration.

Substitute the values in the above expression.

\(\begin{aligned}{l}n &= \frac{{\left( {1000\;{\rm{kg/}}{{\rm{m}}^3}} \right)\left( {\frac{4}{3}\pi {{\left( {0.018\;{\rm{mm}} \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)}^3}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\left( {150\;{\rm{N/C}}} \right)}}\\n \approx 1.0 \times {10^7}\end{aligned}\)

Thus, the number of excess electron charges water droplet must have is the number of excess electrons is \(1.0 \times {10^7}\).